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Unformatted text preview: CHEMISTRY 1AO3 TUTORIAL #11 November 22-26, 2010 Solutions: Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Page 1 of 10 The following problems (1-10) are based on material from Chapter 16 and class notes. Assume all calculations are at 25C, so that K w = 1.0 10-14 (and pK w = 14.00). 1. What is the conjugate acid of each of the following species? Base Conjugate acid (a) S 2 H S (b) CH 3 COO C H 3 COOH (c) ClO 2 H C l O 2 (d) SO 3 2 H S O 3 2. What is the conjugate base of each of the following species? A c i d C o n j u g a t e b a s e (a) H 2 CO 3 H C O 3 (b) HNO 3 N O 3 (c) H 2 PO 4 H P O 4 2 (d) [(CH 3 ) 2 NH 2 ] + (dimethylammonium ion) (CH 3 ) 2 NH (dimethylamine) 3. Write balanced chemical reactions and K a or K b expressions for the first ionization of the following acids and bases ( i.e . assume in each case that only one H + is ionized or added). (a) HF (b) H 3 PO 4 (c) CH 3 NH 2 (d) CH 3 COOH (e) HClO 4 (a) HF + H 2 O F + H 3 O + K a = [F ][ H 3 O + ] [HF] (b) H 3 PO 4 + H 2 O H 2 PO 4 + H 3 O + K a = [H 2 PO 4 ][ H 3 O + ] [H 3 PO 4 ] (c) CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH K b = [CH 3 NH 3 + ] [OH ] [ C H 3 NH 2 ] (d) CH 3 COOH + H 2 O H 3 O + + CH 3 COO K a = [H 3 O + ] [CH 3 COO ] [ C H 3 COOH] (e) HClO 4 is a strong acid and is fully dissociated at all concentrations in water. The equilibrium constant is much, much greater than 1, so we normally do not write it. CHEMISTRY 1AO3 TUTORIAL #11 November 22-26, 2010 Solutions: Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Page 2 of 10 4. (a) What is the pH of a 0.035 M solution of Ba(OH) 2 (aq)? (b) What is the pH of a 0.0042 M solution of HClO 4 (aq)? (c) What is the pOH of the solution in part (b)? (d) A 30.0 mL sample of HClO 4 was diluted to 500 mL, and the pH of the final solution was 4.56. What was the concentration of HClO 4 in the original sample? SOLUTION: (a) Ba(OH) 2 (aq) is a soluble hydroxide, and a strong base. It dissociates fully in water, so Ba(OH) 2 (aq) produces Ba 2+ (aq) and 2 OH (aq). The original 0.035 M sample of Ba(OH) 2 (aq) therefore produces 0.035 M Ba 2+ ions, and 2 (0.035M) OH ions (according to the formula of our hydroxide). [ OH] = 2(0.035M) = 0.070 M From this we can find pOH: pOH = -log[ OH] = -log[0.070] = 1.15 From pH + pOH = 14.00 , rearrange to get (2 sig figs. becomes 2 dec. places) pH = 14.00 pOH = 14.00 1.15 = 12.85 (b) HClO 4 is a strong acid, and dissociates fully in water to give: HClO 4 (aq) + H 2 O(l) ClO 4 (aq) + H 3 O + (aq) The original 0.0042 M sample of HClO 4 (aq) therefore produces 0.0042 M ClO 4 ions, and 0.0042 M H 3 O + ions pH = -log[H 3 O + ] = -log(0.0042) = 2.38 (c) pOH = 14.00 pOH = 14.00 2.38 = 11.62 (d) Since HClO 4 is a strong acid, it dissociates fully in solution. In the final solution, is a strong acid, it dissociates fully in solution....
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This note was uploaded on 12/15/2010 for the course CHEM 1a03 taught by Professor Landry during the Fall '08 term at McMaster University.
- Fall '08